Find the Inverse y=(2x)/5+2

$y = \frac{2 x}{5} + 2$

Interchange the variables.

$x = \frac{2 y}{5} + 2$

Solve for $y$ .

Rewrite the equation as $\frac{2 y}{5} + 2 = x$ .

$\frac{2 y}{5} + 2 = x$

Subtract $2$ from both sides of the equation.

$\frac{2 y}{5} = x - 2$

Multiply both sides of the equation by $\frac{5}{2}$ .

$\frac{5}{2} \cdot \frac{2 y}{5} = \frac{5}{2} (x - 2)$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{5}{2} \cdot \frac{2 y}{5}$ .

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5}{2} \cdot \frac{2 y}{5} = \frac{5}{2} (x - 2)$

Rewrite the expression.

$\frac{1}{2} (2 y) = \frac{5}{2} (x - 2)$

Cancel the common factor of $2$ .

Factor $2$ out of $2 y$ .

$\frac{1}{2} (2 (y)) = \frac{5}{2} (x - 2)$

Cancel the common factor.

$\frac{1}{2} (2 y) = \frac{5}{2} (x - 2)$

Rewrite the expression.

$y = \frac{5}{2} (x - 2)$

Simplify the right side.

Simplify $\frac{5}{2} (x - 2)$ .

Apply the distributive property.

$y = \frac{5}{2} x + \frac{5}{2} \cdot - 2$

Combine $\frac{5}{2}$ and $x$ .

$y = \frac{5 x}{2} + \frac{5}{2} \cdot - 2$

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

$y = \frac{5 x}{2} + \frac{5}{2} \cdot (2 (- 1))$

Cancel the common factor.

$y = \frac{5 x}{2} + \frac{5}{2} \cdot (2 \cdot - 1)$

Rewrite the expression.

$y = \frac{5 x}{2} + 5 \cdot - 1$

Multiply $5$ by $- 1$ .

$y = \frac{5 x}{2} - 5$

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = \frac{5 x}{2} - 5$

Verify if $f^{- 1} (x) = \frac{5 x}{2} - 5$ is the inverse of $f (x) = \frac{2 x}{5} + 2$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (\frac{2 x}{5} + 2)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{2 x}{5} + 2) = \frac{5 (\frac{2 x}{5} + 2)}{2} - 5$

Simplify each term.

Cancel the common factor of $\frac{2 x}{5} + 2$ and $2$ .

Factor $2$ out of $5 (\frac{2 x}{5} + 2)$ .

$f^{- 1} (\frac{2 x}{5} + 2) = \frac{2 (5 (\frac{x}{5} + 1))}{2} - 5$

Cancel the common factors.

Factor $2$ out of $2$ .

$f^{- 1} (\frac{2 x}{5} + 2) = \frac{2 (5 (\frac{x}{5} + 1))}{2 (1)} - 5$

Cancel the common factor.

$f^{- 1} (\frac{2 x}{5} + 2) = \frac{2 (5 (\frac{x}{5} + 1))}{2 \cdot 1} - 5$

Rewrite the expression.

$f^{- 1} (\frac{2 x}{5} + 2) = \frac{5 (\frac{x}{5} + 1)}{1} - 5$

Divide $5 (\frac{x}{5} + 1)$ by $1$ .

$f^{- 1} (\frac{2 x}{5} + 2) = 5 (\frac{x}{5} + 1) - 5$

Apply the distributive property.

$f^{- 1} (\frac{2 x}{5} + 2) = 5 (\frac{x}{5}) + 5 \cdot 1 - 5$

Cancel the common factor of $5$ .

Cancel the common factor.

$f^{- 1} (\frac{2 x}{5} + 2) = 5 (\frac{x}{5}) + 5 \cdot 1 - 5$

Rewrite the expression.

$f^{- 1} (\frac{2 x}{5} + 2) = x + 5 \cdot 1 - 5$

Multiply $5$ by $1$ .

$f^{- 1} (\frac{2 x}{5} + 2) = x + 5 - 5$

Combine the opposite terms in $x + 5 - 5$ .

Subtract $5$ from $5$ .

$f^{- 1} (\frac{2 x}{5} + 2) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\frac{2 x}{5} + 2) = x$

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (\frac{5 x}{2} - 5)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{5 x}{2} - 5) = \frac{2 (\frac{5 x}{2} - 5)}{5} + 2$

Simplify each term.

Cancel the common factor of $\frac{5 x}{2} - 5$ and $5$ .

Factor $5$ out of $2 (\frac{5 x}{2} - 5)$ .

$f (\frac{5 x}{2} - 5) = \frac{5 (2 (\frac{x}{2} - 1))}{5} + 2$

Cancel the common factors.

Factor $5$ out of $5$ .

$f (\frac{5 x}{2} - 5) = \frac{5 (2 (\frac{x}{2} - 1))}{5 (1)} + 2$

Cancel the common factor.

$f (\frac{5 x}{2} - 5) = \frac{5 (2 (\frac{x}{2} - 1))}{5 \cdot 1} + 2$

Rewrite the expression.

$f (\frac{5 x}{2} - 5) = \frac{2 (\frac{x}{2} - 1)}{1} + 2$

Divide $2 (\frac{x}{2} - 1)$ by $1$ .

$f (\frac{5 x}{2} - 5) = 2 (\frac{x}{2} - 1) + 2$

Apply the distributive property.

$f (\frac{5 x}{2} - 5) = 2 (\frac{x}{2}) + 2 \cdot - 1 + 2$

Cancel the common factor of $2$ .

Cancel the common factor.

$f (\frac{5 x}{2} - 5) = 2 (\frac{x}{2}) + 2 \cdot - 1 + 2$

Rewrite the expression.

$f (\frac{5 x}{2} - 5) = x + 2 \cdot - 1 + 2$

Multiply $2$ by $- 1$ .

$f (\frac{5 x}{2} - 5) = x - 2 + 2$

Combine the opposite terms in $x - 2 + 2$ .

Add $- 2$ and $2$ .

$f (\frac{5 x}{2} - 5) = x + 0$

Add $x$ and $0$ .

$f (\frac{5 x}{2} - 5) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{5 x}{2} - 5$ is the inverse of $f (x) = \frac{2 x}{5} + 2$ .

$f^{- 1} (x) = \frac{5 x}{2} - 5$

Find the Inverse y=(2x)/5+2

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